## Real MATHEMATICS – Geometry #20

Escape From Alcatraz

Imagine a classroom that has 5 meters between its walls in length. Tie a 6-meter long rope between these walls. Let the rope be 2 cm high off the ground. Since the rope strained to its limits, its 1-meter long part hangs from either side of the rope.

The ultimate goal is to escape from the classroom from under this rope, without touching the rope.

Rules

• Escape should be from the middle point of the rope.
• One should use the extra part of the rope to extend it.
• One of the students will help you during the escape. He/she will strain the rope for you so that you can avoid touching the rope.
• Each student has exactly one try for his/her escape.

Winning Condition: Using the least amount of rope for your escape.

Football Field

Legal-size for a football field is between 90 and 120 meters in length. Assume that we strain a rope on a football field that is 100 meters long. We fixed this rope right in the middle of both goals while the rope is touching the pitch.

The middle of the rope sits right on the starting point of the field. This is also known as the kick-off point.

Let us add 1 meter to the existing rope. Now, the rope sits flexed, not strained, on the field.

Question: If we try to pick the rope up at the kick-off point, how high will the rope go?

Solution

We can express the question also as follows:

“Two ropes which have length 100m and 101m are tied between two points sitting 100m apart from each other. One picks the 101m-long rope up from its middle point. How high the rope can go?”

If we examine the situation carefully, we can realize that there are two equal right-angled triangles in the drawing:

Using Pythagorean Theorem, we can find the length h:

(50,5)2 = 502 + h2

h ≈ 7,089 meters.

Conclusion

Adding only 1 meter to a 100-meter long rope helps the rope to go as high as 7 meters in its middle point. This means that a 1-meter addition could let an 18-wheeler truck pass under the rope with ease.

M. Serkan Kalaycıoğlu

## Real MATHEMATICS – Geometry #19

How Much Chocolate?

It is midnight and my stomach is talking to me. I hope to find something to eat in the kitchen and I see a chocolate bar:

Immediately made myself a cup of coffee and broke a piece of the chocolate bar:

After I “killed” the broken piece I started having second thoughts about my decision: Oh God; Did I eat too much chocolate?

I placed the leftover on a grid. This way I found where both whole and the broken piece lies on the grid:

The broken piece is shaped like a simple polygon. My goal is to calculate the area of that piece. There are several ways I could calculate the area. Although, the first thing comes to my mind is a theorem called “Gauss’ shoelace theorem”.

Gauss’ Shoelace Theorem

The shoelace theorem can only be applied to simple polygons. In order to use the theorem, I have to find where the edges of the simple polygon lie on the grid:

Theorem uses these points just like shoelaces. But first we have to define the edges and make a list of them:

Do not forget to add the first edge to the bottom of the list.

Now you can multiply the numbers diagonally; from right to left and left to right. Then add left to right and subtract it from right to left ones:

This slideshow requires JavaScript.

{(0*0) + (5*1) + (4*3) + (5*3) + (0*0)} – {(0*5) + (0*4) + (1*5) + (3*0) + (3*0)}

{32} – {5}

27.

The area of that simple polygon can be found by dividing the result above:

27/2

13,5

M. Serkan Kalaycıoğlu

## Real MATHEMATICS – Geometry #18

Protect the Cake

Holy cake is coming to the school today. It will be open for visit in one of the classrooms. You are responsible for the protection of the holy cake. But you have limited people under your command which is why you should use minimum number of guards to protect the holy cake.

About the guards: A museum guard stays at a fixed point and observes the art piece(s) from that point. Obviously guard can see every angle around him/herself by simply rotating.

Floor plan of the classroom is like the following:

Find the minimum number of guards needed in order to protect the holy cake in this room.

Polygonal Rooms

First we will be examining the simplest polygon.

Example 1: Triangle room.

Assume that holy cake is inside a triangle room. One guard is enough for protecting the cake as he/she will be able to see every part of the room:

Example 2: Rectangle room.

Again one guard will be enough:

Q: Is one guard enough for all polygon-shaped rooms?

Convex-Concave: A polygon is convex if each of its internal angles is less than 180 degrees. Otherwise that polygon is called concave.

As we can see above, all convex polygons can be protected by a single guard. Although it can’t be said for all concave polygons.

Now we know that the room of holy cake is a concave polygon. Let’s examine a simple concave polygon first:

It is possible to protect the room if we place the guard right on the vertex of the angle that is greater than 180 degrees:

If we have a room plan as following:

In such room, one guard doesn’t give enough coverage:

Art Gallery Problem

Before moving on the more complex room plans, we have to check if there is any algorithm for finding the necessary number of guards in concave polygons. This problem was first posed by a mathematician named Victor Klee in 1973 and is called art gallery problem. In order to find the necessary number of guards in a room, we should use a method called triangulation.

Triangulation: Basically it is a process that enables us to divide any polygon into triangles.

Let’s take a concave polygon as an example. First we should use triangulation:

Then we should color the vertices of the triangles. In any triangle vertices must have different colors:

Number of guards is the least used vertex color. If guards take positions at those vertices, they will be able to see the room completely:

Least numbered colors are 2 and 3. Placing guards on either of them is enough to cover the whole room. If, let’s say, we choose to place two guards on the color 3, room will be protected.

Solution

We are finally ready to solve the holy cake problem. First we should triangulate the floor plan of the classroom:

Next, we should color the vertices of the triangles:

Three colors were used as follows:

The least used colors are 1 and 3. This gives us the number of guards needed (that is 6) in order to protect the holy cake. If, let’s say we choose to place the guards on the vertices with color 1, room will be covered completely.

One wonders…

1. What would the answer be if there were columns inside the classroom as shown below?
2. Can you find a relationship between the number of vertices of the polygon and number of guards?

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #17

Why do we have round wheels in cars and bicycles?

Square Wheels

Let’s construct a square wheel like the following and test if it can be used as an efficient wheel:

After turning the squared-wheel 45 degrees the square will look like the one on the right:

At this position it is clear even to the naked eye that the height of the wheel is taller comparing to its original state. If we continue turning the wheel for another 45 degrees it goes back to its original position.

Squared-wheel has big disadvantages. A car or a bicycle with squared-wheels will be doomed as height of the vehicle constantly changes.

Triangle Wheels

Among the triangles equilateral is the best one for constructing a wheel as all of its sides have equal length:

After turning the triangular-wheel 60 degrees to the left, it will have the same height:

Although let’s go 30 degrees back and examine the height of the wheel:

Here height of the wheel is clearly taller than the height of the original state. This proves that it is not appropriate to construct a triangular-wheel on a vehicle. Otherwise you might have problems with your spine.

Power of the Circle

The reason why circle is the most powerful and convenient shape for our vehicles is that neither of its height nor width changes while rotating. This separates circle from all the other polygons and makes it the best shape for wheels.

Nonetheless, one wonders if circle is the only possible shape for wheels.

Reuleaux Triangle

Leonardo da Vinci is one of those names that appear in your mind when someone mentions Renaissance. The relationship between this magnificent figure and Reuleaux triangle comes from a world map that was found inside his pupil Francesco Melzi’s notes:

This is known as one of the very first maps that included America. It is believe that this 1514-dated drawing was due to Leonardo da Vinci. If this is true, then it is safe to assume that da Vinci was the first person ever who used the Reuleaux triangle.

It was Leonhard Euler whom discovered the shape and explained it mathematically almost 200 years after da Vinci’s map. You probably realized this from my articles: “It is either Euler or Gauss.”

How come it is called Reuleaux? Because a century after Euler, a German engineer named Franz Reuleaux discovered a machine using Reuleaux triangle. In 1861 Franz Reuleaux wrote a book that made him famous and today he is known as the father of kinematics.

How to Construct Reuleaux Triangle?

I will show you my favorite method for its construction using three identical circles. First draw a circle that has radius r:

Then pick a point on that circle as a center and draw a second circle again with radius r:

At last, pick one of the crossings of the circles as center and draw a third circle with radius r:

The central area of these three circles is a Reuleaux triangle:

When a Reuleaux triangle is rotating, it will have the same height at all times just like the circle:

One wonders…

1. Using Euclid’s tools (a compass and an unmarked ruler) draw an equilateral triangle.
2. Try to construct Reuleaux triangle with on the equilateral triangle.
3. Can you contruct Reuleaux polygon(s) other than the triangle?

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #16

Smell of a Cake

I smell something wonderful. I beam myself up to the kitchen to investigate the source of this smell. I find it: My mother’s chocolate-chipped cake. As I leaned towards the cake, someone grabs my arm: Mom caught me…

I use the emotional card. She doesn’t buy it anymore. My opponent is experienced; my opponent is winning the battle!

As I was thinking of giving up, she offers me a deal. If I can cut three equal pieces out of this cake, one of the pieces will be mine.

Mom’s conditions:

• Only instrument of measurement allowed for the cut is a compass.
• Goal is to cut three pieces that have the same area. Size of the pieces is up to my cutting skills.
• While making the cut, small differences (as if one area is 3,04 and other is 3,09) will be ignored by the mother.
• Most important condition: Pieces must be in the shape of a ring.
• You only have one chance for cutting. There is no turning back after the knife touches the cake.

Art of Cutting Cakes

I tried to find a method on paper because I satisfy all the conditions for the cut.

First I drew a circle that has center O:

Then I created a chord which is as long as the radius of the circle:

I placed the chord on random places inside the circle and marked chord’s midpoints:

I chose any of those marked points and drew a new circle that has center O:

Then I followed the same procedure inside the new circle:

And finally I did the same things for the third time:

Areas which I colored with pens are equal to each other:

For those who wonder the areas, you could calculate and see the approximate results.

One wonders…

I found out that it is possible to cut equal areas that are ring-shaped with using only a compass as mother asks.

Now think: How long the chord should be in order to cut the biggest possible piece?

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #15

Drawing a square

I am dealing with geometry and I imagine that I am in ancient Greece again. Aegean sea is in front of me and I am sitting on a marble between two huge white columns while holding an unmarked ruler and a compass.

First I draw a circle that has center at A and has radius r:

Then I draw the same circle but taking its center at B this time:

I connect the points A and B with a straight line. Then I draw two perpendiculars from the endpoints of the line AB:

I connect the point E to the point F and end up with the ABEF square which has side lengths r:

Biggest circle that can be drawn into ABEF will have diameter r and touch the square at exactly four points:

Area

In order to find the area of a square one can take the square of one side that gives r2.

To find a circle’s area one should multiply the square of the radius with π. In our inscribed circle we calculate the area as πr2/4.

Ratio of these areas would give π/4.

Weight

Now let’s make an experiment. For that all you need is some kind of cardboard cut as a square and a precision scale. Using the scale find the weight of the square-shaped cardboard.

Then draw the biggest possible circle inside this square. Cut that circle out and find its weight with the scale.

Since we are using the same material ratio of the weights should be equal to the ratio of the areas. From here one can easily find an approximation for the number π:

0,76/0,97 = π/4

3,1340… = π

One of the main reasons why we only found an approximation is that the cardboard might not be homogeneous. In other words the cardboard might not have equal amount of material on every point of itself.

Another reason for finding an approximation is that I didn’t cut the square and the circle perfectly.

One wonders…

Draw a circle and then draw the biggest-possible square inside that circle. Find their areas and measure their weights. See if you found an approximation.

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #6

Mathematics was crucial for mankind before ancient Greeks came along. Humans needed mathematics to solve their everyday problems and that is why they were learning it. But ancient Greeks changed that as they developed mathematics for joy. This is one of the reasons why they didn’t limit themselves to the daily problems.

One of the problems ancient Greeks considered is today known as the Delos Problem, or Doubling the Cube. Even the brightest philosophers were helpless against this specific problem. Now I will tell you two common told stories about how Greeks started dealing with this problem.

Surviving the Plague

According to Theon of İzmir (a city in modern Turkey), this story was inside one of the books of Eratosthenes that were lost.

Around 430 BC a devastating plague had arisen in ancient Athens. Leaders of the city were desperate against the plague and they had no idea how to save the people of Athens. During the plague God speaks to the people through an oracle: In order to stop the plague they had to build a new altar. But this altar should have twice the volume of the previous altar.

It was seemingly an easy task for the engineers of the Athens. Although they were unable to build the altar as God wanted them to. According to Plato, Greeks were in illusion as they claimed to know everything about geometry. And with this task God was teaching them a lesson. Plato thought God didn’t want people to build the altar. He only wanted to show people how ignorant they are.

Grave of Glaucus

Second story is being told in one of Archimedes’ books. Apparently Eratosthenes wrote a letter to the King of Greece and mentioned this story.

Zeus and Europa had a son named Minos. King Minos is one of the leading characters in the Greek mythology. In the story it is being told that King Minos’ son Glaucus died at an early age. King wanted his engineers to build a massive grave for his late son. Eventually King thought the grave that was built was rubbish and wasn’t suitable for a royal. He ordered his engineers to double the volume of the cube-shaped grave. In order to do that Minos told the engineers to double the sizes of the grave.

This caused a huge problem as new volume turned out to be eight times the old volume when the sides of the cube-shaped grave were doubled. Neither Minos nor his men were unable to solve this problem.

Three Impossible Problems

I have to remind you that these men had only a compass and an unmarked ruler when they were dealing with this problem. But little they knew was that doubling the cube was one of the three problems that can’t be solved with a compass and an unmarked ruler. (I’ll be talking about the other two in the upcoming articles.) Gauss was the first person who claimed this but he didn’t back his claims with a proof. The first proof came from Pierre Wantzel in 1837! It means at least 2250 years after the problem first came out.

Let’s try to solve the problem with modern mathematics notations:

Assume that we have a cube that has 1 unit sides. Its volume is 1*1*1=1 unit. Doubling the volume of a cube makes 2 units of volume. Then we must find the cube that has volume 2. If such cube has sides a, volume of that cube become a*a*a = a3.

Thus,

a3 = 2

a = 3√2.

We solved the unsolvable… or did we?

Obviously we managed the solve it. But ancient Greeks didn’t have our modern mathematics notations. Actually they didn’t even have numbers. They had to find 3√2 length with an unmarked ruler and a compass. Even with our marked rulers, it is impossible to find how long 3√2 is.

How Long?

In order to find how long 3√2 is, we can use a method called Neusis Drawing. But I will use the power of origami and show you how to find that irrational length.

First of all I took a square paper and using origami techniques to divide the square into three equal parts.

Then I folded the paper such that point A touches the left side of the square as point B touches the line that is in the height of point C.

I called the point A touched on the left side as D. Distance from D to F is 3√2 times the distance from D to E.

Here is how Peter Messer showed this origami technique:

One wonders…

A question that was keeping even the most brilliant minds busy for more than 2000 years can be solved in the matter of seconds using origami. How can this happen? What is the missing sides of compass and ruler?

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #5

I was wondering; if there was a list of hall of fame for famous ancient Greeks Pythagoras would find himself in the top ten for sure. What is striking about his fame is that it comes directly from a geometry property. Although mathematicians know that so called Pythagorean Theorem was known to other cultures at least 1000 years before he “discovered” it.

Pythagorean Theorem: In a right-angled triangle sum of the squares of the perpendicular sides gives the square of the hypotenuse that is the longest side of the triangle.

It is being told that there are 367 different proofs for this theorem. Some of them are so similar, even mathematicians have trouble seeing the difference among these proofs.

Let’s check a few of the proofs.

Proof 1

Elisha Loomis talks about a proof for the Pythagorean Theorem in his book “The Pythagorean Proposition”. This proof is special because it came from a high school student named Maurice Laisnez.

I decided to use cutting papers for the explanation. First of all I cut a random right-angled triangle and then made 3 more copies of it.

I lined these four triangles up such that it gave me a square inside a square:

Since sides of the inner square are c, it has area c2.

Now let’s line the triangle as follows:

Marked areas 1 and 2 are squares and their area is equal to the area of the inner square from the previous alignment. Now let’s find the areas of 1 and 2: They make a2 and b2.

Their addition will make c2. Hence:

a2 + b2 = c2

Proof 2

For the second proof I decided to go to the ancient China.

Zhoubi Suanjing is believed to be written around 500 BC to 200 BC. In the Loomis’ book you can find this proof in the page 253.

Pythagorean Theorem’s proof in the Suanjing.

Again I will cut four right-angled triangles for the explanation of the proof. But this time I will cut the triangles such that their perpendicular sides will have length 3 and 4 units. Chinese mathematicians tried to find the third side of the triangle as follows.

In order to start the proof I lined the triangles up like below and a tiny square formed in the middle:

Tiny square A has sides that have 1 unit each. This is why area of A is 1 unit as well.

We know that the area of one triangle is (3*4)/2 = 6 units. There are four of such triangles and that gives us 6*4 = 24 units of area. When I add the area of A to this result, I can find area of the whole square as 25 units.

If area of a square is 25 units, its one side is square root of the area: √25 = 5 units.

From here we found length of the third side from the triangles:

This proof shows us that 3-4-5 triangle and Pythagorean Theorem were both known in ancient China.

One wonders…

A farmer dad wants to retire. He would like to divide three of his lands to his two sons equally. But he wants to do that without dividing the lands from each other. What should he do?

X-Y-Z are squares as DCG is a right triangle.

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #4

Geometry has a sacred book: Elements. Author of Elements, Euclid, used only these tools when he discovered his geometry:

A compass and an unmarked ruler… He showed what can or can’t be done with them in geometry.

Dividing a Finite Straight Line into Two Equal Parts

We already know that one can draw a straight line segment between any two points. Let’s say we have a straight line between the points A and B. Euclid found an ingenious method to divide AB into two equal parts using his only two tools. Let’s assume that AB is 6 cm.

This way we will know that Euclid’s method works only if we find two parts that are 3 cm long.

According to Euclid one should take point A and point B as the centers of two equal circles with radius AB:

Euclid says that one should define the intersection points as C and D:

Then he suggests one to connect C to D:

At this point Euclid talks about two outcomes:

1. CD is perpendicular to AB.
2. CD divides AB into two equal parts.

As seen in the picture CD really divides AB into two equal parts. One can use this method and see that those two lines are perpendicular with a quadrant.

Dividing a Random Angle into Two Equal Parts

Obviously Euclid didn’t stop there and tried to figure out if it was possible to divide any given angle into two equal parts. Let’s consider straight lines AB and BC intersects and form a 90-degree angle:

Euclid says that one should take B (the intersection point) as center and draw a circle with random radius. This circle will intersect AB and BC at two points: D and E.

Now Euclid tells us to use our compass and draw two equal circles that have centers D and E:

As seen above, these circles intersect at two points. Let’s choose the point F and connect it with B:

Euclid claims that the straight line BF divides the angle into two equal parts:

We can see with our quadrant angle is divided into two equal parts of 45 degrees.

One wonders…

Is it possible to use Euclid’s methods and divide any given straight line into three equal parts?

Try to answer the same thing for an angle.

M. Serkan Kalaycıoğlu

## Real Mathematics – Geometry #3

“There is no royal road to geometry.”

From Euclid to the king who asked Euclid if there is an easier way to learn geometry.

Up until now I have mentioned Euclid and his book Elements a few times. This masterpiece is actually a collection of 13 books and was considered as the source of only known geometry for thousands of years. Historical figures including Newton, Leibniz, Omar Khayyam and many others learned mathematics through Euclid’s Elements.

First book of Elements starts with 23 seemingly obvious and simple definitions. I will mention some of them below.

Elements Book I

Definition 1: A point is that of which has no parts. (Zero dimensions)

Definition 2: A line is length without breadth. (One dimension)

Definition 3: The extremities of a line are points.

Definition 4: A straight line is any one which lies evenly with points itself.

Definition 8: A plane angle is the inclination of the lines to one another when two lines in a plane meet one another and are not lying in a straight-line.

Definition 15: A circle is a plane figure contained by a single line such that all of the straight-lines radiating towards from one point amongst those lying inside the figure are equal to one another.

After reading these definitions for the first time, a few question marks popped up in my head.

For instance the first definition suggests that a point has no dimensions. If that’s so, how can one show a point lying on a plane?

Is it even possible to show something that has no dimensions?!

Which of these two can suggest a point to us? Obviously their sizes don’t matter and neither of them is an illustration of an actual point.

In this context, second definition is not different from the first one: One can’t draw something that has no breadth.

Eighth definition is about angles. In order to draw an illustration for a random angle one must know how to draw lines, straight lines and dots.

I’ve just showed you that even basic geometrical shapes are impossible to demonstrate. We can only imagine them in our minds. This means that in a way architects are selling illusions.

It is being told that mathematics has abstract and tangible parts. Whenever a student is dealing with abstract mathematics, teacher ought to give tangible examples so that student can comprehend with the subject easily. Nevertheless, we are helpless even when we want to give a full tangible explanation to a simple thing like a straight line.

Magic inside the Elements

In the first proposition of the first book of Elements given a random straight line, Euclid is showing us how to draw an equilateral triangle from that line.

Just to remind you, Euclid only used an unmarked ruler and a compass in his methods. Stop here and try to think of a way to construct an equilateral triangle from a random straight line.

Euclid’s Method

1. Assume that we have a finite straight line AB.
2. Take AB as radius and draw a circle that has center A.
3. Now take AB as radius and draw another circle that has center B this time.
4. These circles will intersect at two points. Call one of them C.
5. Connect A to C. One can easily see that AB and AC are radii; hence they are equal in length.
6. Then connect B to C. One can observe that BC and BA are radii; hence they are equal in length.
7. AB and AC, BA and BC are equal. Since AB and BA are the same straight line one can conclude that AB=AC=BC.
8. These three straight lines construct an equilateral triangle.

One wonders…

These methods are taken from a book that was written around 2300-2400 years ago. What I find fascinating about mathematics is that we are not even capable of showing what a dot is, but we can also explore other planets using the power of the language of mathematics.

Now use Euclid’s materials (an unmarked ruler and a compass) and try to draw the twin of a given random straight line. Hint: Analyze the second proposition of the book I of Elements.

M. Serkan Kalaycıoğlu